In a gravitational field, the net work done by the field when an object moves from rest at point A to rest at point B depends only on the positions of A and B, not the path taken or the object's velocity while moving.
Explanation:When an object moves from rest at point A to rest at point B in a gravitational field, the net work done by the field is dependent on the positions of A and B only (option 5). This is based on the principle of conservation of energy in physics. If we consider the gravitational field to be conservative, the work done is independent of the path taken, and depends only on the initial (A) and final points (B).
The mass of the object and the gravitational field strength will determine the gravitational potential energy at these points, hence affecting the work done when the object is moved. It does not depend on the nature of the external force moving the object or the velocity of the object while moving. These factors could affect the time it takes to move the object, but not the net work done by the gravitational field.
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The net work done by the field when an object is moved from rest at point A to rest at point B in a gravitational field depends on the nature of the external force moving the object from A to B, and both the positions of A and B and the path taken between them.
Explanation:The net work done by the field when an object is moved from rest at point A to rest at point B in a gravitational field depends on the nature of the external force moving the object from A to B, and both the positions of A and B and the path taken between them.
If the object is moved vertically upwards or downwards, the work done by the gravitational field will be positive or negative respectively. However, if the object is moved in a horizontal direction, there will be no work done by the field as there is no change in height.
Therefore, option 3, both the positions of A and B and the path taken between them, is the correct answer.
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In order to understand the full scope of a disease, we take its occurrence into account. The __________ of a disease is the number of people in a population who develop a disease at a specified time.
a. incidence
b. PREVALENCE
c. endemic infection
d. sporadic infection
Answer: b. PREVALENCE
Explanation:
Prevalence can be defined as the proportion of individuals in a population having a disease or common characteristic. Prevalence of a particular disease is number of people in a population who have develop a disease at a particular time.
It can be expressed in form of proportion Which equals the number of developed cases at a particular time divided by the total number of population.
A calorimeter is used to determine the specific heat capacity of a test metal. If the specific heat capacity of water is known, what quantities must be measured?
Answer:
initial and final temperatures of both the water and metal, mass of the metal, and mass of the water
Explanation:
Heat lost by the metal, [tex]Q = mc(t_{2} - t_{1})[/tex]
Heat gained by the water in the calorimeter, [tex]Q_{w} = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
For energy to be conserved in the system, the heat lost by the metal will equal the heat gain by the water in the calorimeter.
[tex]mc(t_{2} - t_{1}) = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
Where,
m is the mass of the metal
c is specific heat capacity of the metal
t₂ is the final temperature of the metal
t₁ is the initial temperature of the metal
[tex]m_{w} [/tex] is the mass of the water
[tex]c_{w} [/tex] is specific heat capacity of water
[tex]t_{2w} [/tex] is the final temperature of water
[tex]t_{1w} [/tex] is the initial temperature of water
From the question given, specific heat capacity of the water is known, the quantities to be measured are;
Initial and final temperatures of both the water and metal,
Mass of the metal, and mass of the water
A projectile of mass 0.607 kg is shot straight up with an initial speed of 20.1 m/s. (a) How high would it go if there were no air resistance?
Answer:
20.6m
Explanation:
The mass is useless in this case.
Kinetic Energy = Potential Energy
g is acceleration due to gravity.
(1/2)mv² = mgh (m's drop out)
(1/2)v² = gh
(1/2)(20.1²) = 9.81(h)
h = (20.1 x 20.1)/9.81 x 2
20.61 meters
Life of a Star Cluster. Imagine you could watch a star cluster from the time of its birth to an age of 13 billion years. Describe in one or two paragraphs what you would see happening during that time?
Explanation:
The life span of a star is quite as many more than 10 billion years. During these many years it would pass through different stages of its life, from nebula to ( perhaps, the current stage) supernova to neutron star or Black Hole.
The stages of the life of star over a span of 13 billion year can be summarized as
the transition stages are
Nebula → Blue star → Blue-white super-giant star → Supernova → Neutron star →Black Hole.
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M, what was the initial concentration?
Answer:
Initial concentration of the reactant = 3.34 × 10^(-2)M
Explanation:
Rate of reaction = 2.30×10−4 M/s,
Time of reaction = 80s
Final concentration = 1.50×10−2 M
Initial concentration = Rate of reaction × Time of reaction + Final concentration
= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M
Initial concentration = 3.34 × 10^(-2)M
A circuit has a current of 2.4 A. The voltage is increased to 4 times its original value, while the resistance stays the same. How should the resistance change to return the current to its original value if the voltage remains at its increased amount?
Answer:
Resistance will become 4 times the previous value
Explanation:
We have given current in the circuit i = 2.4 A
According to ohm's law current in the circuit is given by [tex]I=\frac{V}{R}[/tex]
So [tex]2.4=\frac{V}{R}[/tex]............eqn 1
Now voltage is increased to 4 times so new voltage = 4 V
And current in the circuit is same as 2.4 A
We have to fond the resistance so that after increasing voltage current will be same
So [tex]2.4=\frac{4V}{R_{unknown}}[/tex]..........eqn 2
Dividing eqn 1 and 2
[tex]1=\frac{V}{R}\times \frac{R_{unkown}}{4V}[/tex]
[tex]R_{unknown}=4R[/tex]
So resistance will become 4 times the previous value
Answer:
It should increase to four times its value.
Explanation:
Where did Earth’s water come from?, select the two competing scientific theories supported by evidence that explain the formation of the hydrosphere.
Answer:
Explanation:
There are two prevailing theories: One is that the Earth held onto some water when it formed, as there would have been ice in the nebula of gas and dust (called the proto-solar nebula) that eventually formed the sun and the planets about 4.5 billion years ago. Some of that water has remained with the Earth, and might be recycled through the planet's mantle layer, according to one theory.
The second theory holds that the Earth, Venus, Mars and Mercury would have been close enough to that proto-solar nebula that most of their water would have been vaporized by heat; these planets would have formed with little water in their rocks. In Earth's case, even more water would have been vaporized when the collision that formed the moon happened. In this scenario, instead of being home-grown, the oceans would have been delivered by ice-rich asteroids, called carbonaceous chondrites.
Two theories explain the formation of Earth's hydrosphere: water from interstellar grains and water from comets and asteroids.
Explanation:There are two competing scientific theories supported by evidence that explain the formation of Earth's hydrosphere:
Water from comets and asteroids: Another theory suggests that the water may have been brought to Earth when comets and asteroids impacted it. Scientists estimate that comet impacts during Earth's early years could have contributed enough water to account for what we see today.
Which of these statements best describe a leader? A. A leader is any person who is a part of the management team. B. A leader is any person who is always the first to perform any business activity. C. A leader is a person who inspires other team members to reach the organizational goals D. A leader is any person who puts others’ interest before his or her own. E. A leader is the one who owns the business and runs it on a day to day basis.
Answer:
I think the answer is c
Explanation:
But it is kinda of opinionated
Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals
What are the qualities of leader?A leader is someone who motivates others to achieve the organization's objectives.
A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.
A leader is a person who inspires other team members to reach the organizational goals
Hence,statements C best describe a leader.
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Does air pressure increase or decrease with an increase in altitude?
Answer:
increase
Explanation:
think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing
Answer:
decrease.
Explanation:
The air pressure is given by
P = h x d x g
where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.
As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.20 m/s. Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 43.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 5.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? Where is Henrietta when she catches the bagels?
Answer:
[tex]u_x=8.5454\ m.s^{-1}[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Explanation:
Given:
speed of walking of Henrietta, [tex]v_w=3.2\ m.s^{-1}[/tex]height of projection of projectile, [tex]h=43.9\ m[/tex]Horizontal distance between the window and Henrietta when the projectile was launched:[tex]r=3.2\times 5[/tex]
[tex]r=16\ m[/tex]
Since the projectile was thrown at the time when she had passed below the window 5 seconds ago.Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.
Now the time taken for the object to reach to Henrietta ignoring the height where she catches:
[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]
[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]
[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.
Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:
[tex]\Delta d=v_w\times t[/tex]
[tex]\Delta d=9.5782\ m[/tex]
Now the total horizontal distance from the window to Henrietta when the projectile reached her:
[tex]d=\Delta d+r[/tex]
[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.
Now the initial horizontal velocity of launch of the projectile:
[tex]u_x=\frac{d}{t}[/tex]
[tex]u_x=\frac{25.5782}{2.9932}[/tex]
[tex]u_x=8.5454\ m.s^{-1}[/tex]
What is the mechanical advantage of the lever if you hold on to the end of the long side and lift something that's placed on the short side?
Answer:
[tex]MA=\frac{F_l}{F_e}>1[/tex]
Explanation:
The mechanical advantage with a shorter lever will always be greater than 1.
It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.
[tex]F_e\times d_e=F_l\times d_l[/tex]
where:
[tex]F_e=[/tex] force on the effort arm
[tex]F_l=[/tex] force on the load arm
[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.
So, one factor balances the other keeping the product of the two constant.
And we know that mechanical advantage is :
[tex]MA=\frac{F_l}{F_e}[/tex]
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a uniform electric fiedl of magnitude E.
Answer:
Range, [tex]R = MV²/2QE[/tex]
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
A force of 14 N acts on a 5 kg object for 3 seconds.
a. What is the object’s change in momentum?
b. What is the object’s change in velocity?
Answer: a) 42Nm b) 8.4m/s
Explanation:
Impulse is defined as object change in momentum.
Since Force = mass × acceleration
F = ma
Acceleration is the rate of change in velocity.
F = m(v-u)/t
Cross multiply
Ft = m(v-u)
Since impulse = Ft
and Ft = m(v-u)... (1)
The object change in velocity (v-u) = Ft/m from eqn 1
Going to the question;
a) Impulse = Force (F) × time(t)
Given force = 14N and time = 3seconds
Impulse = 14×3
Impulse = 42Nm
b) The object change in velocity (v-u) = Ft/m where mass = 5kg
v-u = 14×3/5
Change in velocity = 42/5 = 8.4m/s
When you swim in a pool, Select one: a. rolling friction occurs. b. fluid friction occurs. c. sliding friction occurs. d. static friction occurs.
Answer:
b. fluid friction occurs
Explanation:
Water is a liquid and both going through air gas and liquid cause fluid friction.
Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is increasing when the radius is 4.70 in. Assume that π = 3.14.
Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
2-lbm of water at 500 psia intially fill the 1.5-ft3 left chamber of a partitioned system. The right chamber’s volume is also 1.5 ft3, and it is initially evacuated. The partition is now ruptured, and heat is transferred to the water until its temperature is 300°F. Determine the final pressure of water, in psia, and the total internal energy, in Btu, at the final state.
Explanation:
Formula for final volume of chamber if the partition is ruptured will be as follows.
[tex]V_{2}[/tex] = 1.5 + 1.5
= 3.0 [tex]ft^{3}[/tex]
As mass remains constant then the specific volume at this state will be as follows.
[tex]\nu_{2} = \frac{V_{2}}{m}[/tex]
= [tex]\frac{3.0}{2}[/tex]
= 1.5 [tex]ft^{3}/lbm[/tex]
Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.
[tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]
[tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]
[tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]
Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.
[tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]
1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]
[tex]x_{2}[/tex] = 0.229
Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F
and, [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia
Formula to calculate internal energy at the final state is as follows.
[tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]
= [tex]2(269.51 + 0.229 \times 830.45)[/tex]
= 920.56 Btu
Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.
Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would_____________________.
Answer:
Tribe historians would: Depict a significant event for each year on a buffalo or deer skin
Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.
Explanation:Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.
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Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Answer:
374.39 J/K
Explanation:
Entropy: This can be defined as the degree of disorder or randomness of a substance.
The S.I unit of entropy is J/K
ΔS = ΔH/T ..................................... Equation 1
Where ΔS = entropy change, ΔH = Heat change, T = temperature.
ΔH = cm................................... Equation 2
Where,
c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.
Substitute into equation 2
ΔH = 333000×0.3071
ΔH = 102264.3 J.
Also, T = 273.15 K
Substitute into equation 1
ΔS = 102264.3/273.15
ΔS = 374.39 J/K
Thus, The change in entropy = 374.39 J/K
The change in entropy of the gas is 374.39 J/K.
The given parameters:
Mass of the ice, m = 0.3071 kgTemperature of the gas, T = 273.15 KLatent heat of fusion of water, L = 333,000 J/kgThe heat of fusion of the ice is calculated as follows;
[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]
The change in entropy of the gas is calculated as follows;
[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]
Thus, the change in entropy of the gas is 374.39 J/K.
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Complete the following statement:When work is done on a positive test charge by an external force to move it from one location to another, electric potential _________.( increase or decrease).
Answer:
increase
Explanation:
Electric potential is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field. Taking a test positive charge from one point to another means that work is done against the field hence electric potential increases.
If the magnitude of a charge is twice as much as another charge, but the force experienced is the same, then the electric field strength of this charge is _____ the strength of the other charge.
Answer:
The new electric field strength of this charge is half of the strength of the other charge.
Explanation:
The electric force acting on the charge particle is given by :
F = q E
[tex]E=\dfrac{F}{q}[/tex]
Where
q is the charged particle
E is the electric field
If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :
[tex]E'=\dfrac{F}{q'}[/tex]
[tex]E'=\dfrac{F}{(2q)}[/tex]
[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]
[tex]E'=\dfrac{E}{2}[/tex]
So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.
Final answer:
The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.
Explanation:
If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.
If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.
A two stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2. at 10 s after launch the first stage of the rocket (now without fuel) is released. the second stage now had an acceleration of 6 m/s2
a) how high is the rocket when the first stage seperates?
b)how fast is the rocket moving upon first stage seperation?
c) what will be the maximum height attained by the first stage after seperation?
d) what will be the distance between the first and second stages 2 s after separation
Answer:
a) 200m
b) 40 m/s
c) 81.55m
d) 31.62m
Explanation:
Solution
a)
y = y0 + u×t+⅟2×a×t2 =
y0 = 0
u = 0
y = unknown
a = 4m/s2
t = time = 10 seconds
y = 0.5×4×102 = 200m
b) v = u + at
v = 0 + 4×10 = 40 m/s
c) v2 = u2 - 2×g×y
at maximum height v = 0
we have
402 = 2×9.81×y
Y =81.55m
d)
for the stage 2 we haace
y = y0 + u×t+⅟2×a×t2 =
y = 0 + 4×2+0.5×6×22 = 92m
for the stage one we have
y = 0+40×2-0.5×9.81×4= 60.38m
distance between the first and second stage 2s aftee separation = 92-60.38 = 31.62m
John weighs 90 lbs and Jane weighs 60 lbs. They are both sitting on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?
Answer: Jane 6ft from fulcrum
John 4 ft from fulcrum
Explanation: You use the law of moment of force.
That is Clockwise moment equals anti clockwise moment.
Here is the attachment involving the diagram and procedure of calculations and the answers.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
A 130 g ball and a 230 g ball are connected by a 34-cm-long, massless, rigid rod. The balls rotate about their center of mass at 120 rpm .
What is the speed of the 100 g ball?
The linear velocity of the 100 g ball in the rotating system with a 130 g ball and a 230 g ball, connected by a rigid rod and rotating at 120 RPM, is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
Given values:
- Mass of the 130 g ball [tex](\(m_1\))[/tex]: 130 g
- Mass of the 230 g ball [tex](\(m_2\))[/tex]: 230 g
- Length of the rod [tex](\(r\)):[/tex] 34 cm = 0.34 m
- Initial angular velocity [tex](\(ω_{\text{initial}}\))[/tex]: [tex]\( ω_{\text{initial}} = \frac{2π \times 120}{60} \)[/tex]
Now, let's calculate the initial angular velocity:
[tex]\[ ω_{\text{initial}} = \frac{2π \times 120}{60} = 4π \, \text{rad/s} \][/tex]
Next, calculate the moment of inertia for each ball:
[tex]\[ I_1 = \frac{2}{5}m_1r^2 = \frac{2}{5} \times 0.13 \times (0.34)^2 \][/tex]
[tex]\[ I_2 = \frac{2}{5}m_2r^2 = \frac{2}{5} \times 0.23 \times (0.34)^2 \][/tex]
Now, calculate the total moment of inertia:
[tex]\[ I_{\text{total}} = I_1 + I_2 \][/tex]
Substitute the values into the conservation of angular momentum equation:
[tex]\[ I_{\text{total}} \times ω_{\text{initial}} = I_{\text{total}} \times ω_{\text{final}} \][/tex]
Solve for [tex]\(ω_{\text{final}}\)[/tex] and then use it to calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball:
[tex]\[ v = ω_{\text{final}} \times r \][/tex]
After performing these calculations, we can determine the speed of the 100 g ball. Let me provide you with the numerical results in the next response.
After performing the calculations, the final angular velocity [tex](\(ω_{\text{final}}\))[/tex] is determined to be the same as the initial angular velocity [tex](\(4π \, \text{rad/s}\))[/tex]. Now, we can calculate the linear velocity [tex](\(v\))[/tex] of the 100 g ball using the formula [tex]\(v = ω_{\text{final}} \times r\).[/tex]
[tex]\[ v = 4π \times 0.34 \][/tex]
[tex]\[ v \approx 4 \times 3.14 \times 0.34 \][/tex]
[tex]\[ v \approx 4.24 \, \text{m/s} \][/tex]
Therefore, the speed of the 100 g ball in this rotating system is approximately [tex]\(4.24 \, \text{m/s}\).[/tex]
A woman is reported to have fallen 161 ft from a building, landing on a metal ventilator box, which she crushed to a depth of 29 in. She suffered only minor injuries. Ignoring air resistance, calculate:(a) the speed of the woman just before she collided with the ventilator and (b) her average acceleration while in contact with the box. (c) Modeling her acceleration as constant, calculate the time interval it took to crush the box.
Answer:
a) 31.02 m/s
b) - 653.24 m/s²
c) 50 ms
Explanation:
a) For a vertical movement, let's suppose that the woman fell by the rest, thus she's initial velocity v0 = 0 m/s. If she makes a distance (S) of 161 ft = 49.1 m, thus, her final velocity at the box, v is:
v² = v0² +2aS
The acceleration is the gravity acceleration, 9.8 m/s².
v² = 2*9.8*49.1
v² = 962.36
v =√962.36
v = 31.02 m/s
b) She crushes the box until she stops, so her final velocity will be 0 m/s. The initial velocity is now 31.02 m/s, and S = 29 in = 0.7366 m
0 = 31.02² + 2a*0.7366
-1.4732a = 962.36
a = - 653.24 m/s² (the negative signal indicates that she's dessacelaranting)
c) The time can be calculate by:
v = v0 + at
0 = 31.02 -653.24*t
653.24t = 31.02
t = 0.05 s = 50 ms
A spectral line that appears at a wavelength of 321 nm in the laboratory appears at a wavelength of 328 nm in the spectrum of a distant object. We say that the object's spectrum is:
Answer:
We say that the object's spectrum is shifted.
Explanation:
Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
That shift can be used to find the velocity of the object by means of the Doppler velocity.
[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)
Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.
The object's spectrum is affected by the Doppler effect due to its motion away from us.
The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.
The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.
This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.
For more such questions on Doppler effect, click on:
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To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by:_________
(a) the worker’s applied force,
(b) the gravitational force on the crate, and
(c) the normal force exerted by the incline on the crate?
(d) What is the total work done on the crate?
Answer:
a. [tex]W_w=313.5\ J[/tex]
b. [tex]W_g=-155.312\ J[/tex]
c. [tex]F_N=222.045\ N[/tex]
d. [tex]W_t=313.5\ J[/tex]
Explanation:
Given:
angle of inclination of the surface,[tex]\theta=25^{\circ}[/tex]mass of the crate, [tex]m=25\ kg[/tex]Force applied along the surface, [tex]F=209\ N[/tex]distance the crate slides after the application of force, [tex]s=1.5\ m[/tex]a.
Work done by the worker who applied the force:
[tex]W_w=F.s\ cos 0^{\circ}[/tex] since the direction of force and the displacement are the same.
[tex]W_w=209\times 1.5[/tex]
[tex]W_w=313.5\ J[/tex]
b.
Work done by the gravitational force:
[tex]W_g=m.g\times h[/tex]
where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:
[tex]h=s\times sin\ \theta[/tex]
[tex]h=1.5\times sin\ 25^{\circ}[/tex]
[tex]h=0.634\ m[/tex]
So, the work done by the gravity:
[tex]W_g=25\times 9.8\times (-0.634)[/tex] ∵direction of force and displacement are opposite.
[tex]W_g=-155.312\ J[/tex]
c.
The normal reaction force on the crate by the inclined surface:
[tex]F_N=m.g.cos\ \theta[/tex]
[tex]F_N=25\times 9.8\times cos\ 25[/tex]
[tex]F_N=222.045\ N[/tex]
d.
Total work done on crate is with respect to the worker: [tex]W_t=313.5\ J[/tex]
A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has burned?
Answer:
Velocity of rocket will be equal to 7.81 m/sec
Explanation:
We have given mass of the rocket [tex]m_1=4kg[/tex]
Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )
Velocity of fuel [tex]v_2[/tex] = 625 m/sec
We have to find the velocity of rocket [tex]v_1[/tex]
From conservation of momentum we know that initial momentum is equal to final momentum '
So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel
So [tex]4\times v_1=625\times 0.05[/tex]
[tex]v_1=7.81m/sec[/tex]
So velocity of rocket will be equal to 7.81 m/sec
Obliquity describes:
a. the roundness of an object
b. the circularity of the orbit
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
d. none of the above
Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).
An electron moving parallel to a uniform electric field increases its speed from 2.0 ×× 1077 m/sm/s to 4.0 ×× 1077 m/sm/s over a distance of 1.3 cmcm.
What is the electric field strength?
Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
An engineer claims that solar cells for the generation of electrical power can never meet our high demand for electrical energy. The engineer points out that such cells____________________.
Answer:
The sentence is following with "are made from mostly silicon material".
Explanation:
Silicon is extracted and refined by the use of electrical energy. The amount of energy required to produce the silicon in a solar cell is greater than the energy that cell will produce during its useful working life.